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\title{\heiti\zihao{2} 习题6.1}
\author{中书君}
\date{\songti 2021年1月13日}

\begin{document}
\maketitle
\section{验证 $y=\frac{x^{2}}{2} \operatorname{sgn} x$ 是 $|x|$ 在 $(-\infty,+\infty)$ 上的一个原函数.}
\textbf{证}\quad
可验证$y^{\prime}=x \cdot \operatorname{sgn} x=|x|$,所以$y=\frac{x^{2}}{2} \operatorname{sgn} x$ $|x|$ 在 $(-\infty,+\infty)$ 上的一个原函数.

\section{求下列不定积分}
\subsection{$\int \frac{(\sqrt{x}-x)^{3}}{x^{2} \sqrt{x}} \mathrm{~d} x$}
\textbf{解}\quad
$x>0$

$\int \frac{(\sqrt{x}-x)^{3}}{x^{2} \sqrt{x}} \mathrm{~d} x=\int \frac{x \sqrt{x}-3 x^{2}+3 x^{2} \sqrt{x}-x^{3}}{x^{2} \sqrt{x}} \mathrm{~d} x=\int \frac{1}{x} \mathrm{~d} x-\int\frac{3}{\sqrt{x}} \mathrm{~d} x+\int 3 \mathrm{~d} x-\int \sqrt{x} \mathrm{~d} x=\ln x-6 x^{\frac{1}{2}}+3 x-\frac{2}{3} x^{\frac{3}{2}}+C$


\subsection{$\int\left(\mathrm{e}^{x}-2 \sin x+2 x \sqrt{x}\right) \mathrm{d} x$}
\textbf{解}\quad
$x>0$

$\int\left(\mathrm{e}^{x}-2 \sin x+2 x \sqrt{x}\right) \mathrm{d} x=\int e^{x} \mathrm{~d} x-2 \int \sin x \mathrm{~d} x+2 \int x^{\frac{3}{2}} \mathrm{~d} x=e^{x}+2 \cos x+\frac{4}{5} x^{\frac{5}{2}}+C$


\subsection{$\int \cos x \cos 2 x \mathrm{~d} x$}
\textbf{解}\quad
$\int \cos x \cos 2 x \mathrm{~d} x=\int \cos 2 x d \sin x=\int\left(1-2 \sin ^{2} x\right) d \sin x=\sin x-\frac{2}{3} \sin ^{3} x+c$


\subsection{$\int \tan ^{2} x \mathrm{~d} x$}
\textbf{解}\quad
$\int \tan ^{2} x \mathrm{~d} x=\int \frac{\sin ^{2} x}{\cos ^{2} x} \mathrm{~d} x=\int \frac{1}{\cos ^{2} x} \mathrm{~d} x-\int \frac{\cos ^{2} x}{\cos ^{2} x} \mathrm{~d} x=\tan x-x+C$


\subsection{$\int \frac{x^{4}}{x^{2}+1} \mathrm{~d} x$;}
\textbf{解}\quad
$\int \frac{x^{4}}{x^{2}+1} \mathrm{~d} x=\int\left(x^{2}-1\right) \mathrm{~d} x+\int \frac{1}{x^{2}+1} \mathrm{~d} x=\frac{1}{3} x^{3}-x+\arctan x+C$


\subsection{$\int \frac{\cos 2 x}{\cos ^{2} x \cdot \sin ^{2} x} \mathrm{~d} x$}
\textbf{解}\quad
$\int \frac{\cos 2 x}{\cos ^{2}+\sin ^{2} x} \mathrm{~d} x=\int \frac{\cos ^{2} x-\sin ^{2} x}{\cos ^{2} x \sin ^{2} x} \mathrm{~d} x=\int \frac{1}{\sin ^{2} x}-\frac{1}{\cos ^{2} x} \mathrm{~d} x=-\cot x-\tan x +C$


\subsection{$\int\left(1-x^{3}\right)^{2} \mathrm{~d} x$}
\textbf{解}\quad
$\int\left(1-x^{3}\right)^{2} \mathrm{~d} x=\int 1-2 x^{3}+x^{6} \mathrm{~d} x=x-\frac{1}{2}x^4+\frac{1}{7}x^7+C$


\subsection{$\int \sec x(\sec x+\tan x) \mathrm{d} x$}
\textbf{解}\quad
$\int \sec x(\sec x+\tan x) \mathrm{~d} x=\int \frac{\frac{1+\sin x}{\cos x}}{\cos x} \mathrm{~d} x=\int \frac{1+\sin x}{\cos ^{2} x} \mathrm{~d} x=\int \frac{1}{\cos ^{2} x} \mathrm{~d} x-\int\frac{d \cos x}{\cos ^{2} x}=\arctan x+\frac{1}{\cos x}+C$


\subsection{$\int \frac{3^{x}-2^{x}}{6^{x}} \mathrm{~d} x$}
\textbf{解}\quad
$\int \frac{3^{x}-2^{x}}{6^{x}}=\int \frac{1}{2^{x}}-\frac{1}{3^{x}} \mathrm{~d} x=\frac{1}{3^{x} \ln 3}-\frac{1}{2^{x} \ln 2}+C$


\subsection{ $\int \frac{\mathrm{e}^{3 x}-1}{\mathrm{e}^{x}-1} \mathrm{~d} x$}
\textbf{解}\quad
$\int \frac{e^{3 x}-1}{e^{x}-1} \mathrm{~d} x=\int e^{2 x}+2 e^{x}+1 \mathrm{~d} x=\frac{e^{2 x}}{2}+2 e^{x}+x+C$


\subsection{ $\int \sqrt[m]{x^{n}} \mathrm{~d} x$}
\textbf{解}\quad
$\int m \sqrt{x^{n}} \mathrm{~d} x=\int x^{\frac{n}{m}} \mathrm{~d} x$$\int \sqrt[m]{x^{n}} \mathrm{~d} x=\left\{\begin{array}{ll}\frac{m}{n+m} x^{\frac{n+m}{m}}+C, & \frac{m}{n} \neq-1 \\ \ln |x|+C, & \frac{m}{n}=-1\end{array}\right.$


\subsection{ $\int|x| \mathrm{d} x$;}
\textbf{解}\quad
由1.得$\int|x| \mathrm{d} x=\frac{x^{2}}{2} \operatorname{sgn} x+C$


\subsection{ $\int \frac{\sin x}{\sqrt{\cos x}} \mathrm{~d} x$}
\textbf{解}\quad
$\int \frac{\sin x}{\sqrt{\cos x}} \mathrm{~d} x=-\int \frac{d \cos x}{\sqrt{\cos x}}=-2 \cos x^{\frac{1}{2}}+C$


\subsection{$\int \frac{1}{\mathrm{e}^{x}-1} \mathrm{~d} x$}
\textbf{解}\quad
$\int \frac{1}{e^{x}-1} \mathrm{~d} x=\int \frac{1}{e^{x}\left(e^{x}-1\right)} d e^{x}=\int \frac{1}{e^{x}-1}-\frac{1}{e^{x}} d e^{x}=\ln \left|\frac{e^{x}-1}{e^{x}}\right|+C$


\subsection{$\int(\cos x-\sin x) \mathrm{e}^{x} \mathrm{~d} x$}
\textbf{解}\quad
$\int(\cos x-\sin x) \mathrm{e}^{x} \mathrm{~d} x=\mathrm{e}^{x} \cos x+C$


\subsection{$\int \sin 2 x \cos 3 x \mathrm{~d} x$}
\textbf{解}\quad
$\int \sin 2 x \cos 3 x \mathrm{~d} x=\frac{1}{2}\left[\int \sin (5 x) \mathrm{~d} x+\int \sin (-x) \mathrm{~d} x\right]=\frac{1}{2}\left[-\frac{1}{5} \cos 5 x+\cos x\right]+C$


\subsection{$\int\left(a_{n} x^{n}+\cdots+a_{1} x+a_{0}\right) \mathrm{d} x$}
\textbf{解}\quad
$\int \sum_{i=0}^{n} a_{i} x^{i} \mathrm{~d} x=\sum_{i=0}^{n} \frac{a_{i} x^{i+1}}{i+1}+C$


\subsection{$\int(x-\sqrt{x})^{3} \mathrm{~d} x$}
\textbf{解}\quad
$\int(x-\sqrt{x})^{3} \mathrm{~d} x=\int x^{3}-3 x^{2} \sqrt{x}+3 x^{2}-x \sqrt{x} \mathrm{~d} x=\frac{1}{4} x^{4}-\frac{6}{7} x^{\frac{7}{2}}+x^{3}-\frac{2}{5} x^{\frac{5}{2}}+C$


\subsection{$\int(1+x) \sqrt{x \sqrt{x \sqrt{x}}} \mathrm{~d} x$}
\textbf{解}\quad
$\int(1+x) \sqrt{x \sqrt{x \sqrt{x}}} \mathrm{~d} x=\int(1+x) \cdot x^{\frac{7}{8}} \mathrm{~d} x=\frac{8}{15} x^{\frac{15}{8}}+\frac{8}{23} x^{\frac{23}{8}}+C$


\subsection{$\int \frac{1}{1+\cos 2 x} \mathrm{~d} x$}
\textbf{解}\quad
$\int \frac{1}{1+\cos 2 x} \mathrm{~d} x=\int \frac{1}{2 \cos ^{2} x} \mathrm{~d} x=\frac{1}{2} \tan x+C$

\end{document}